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Electric field of infinitely long parallel wires

  1. Oct 26, 2008 #1

    E&M

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    1. The problem statement, all variables and given/known data
    Two infinitely long parallel wires separated by a distance 2d, one carries uniform linear charge density of [tex]\lambda[/tex] and the other one carries an uniform linear charge density of -[tex]\lambda[/tex], find the electric field at a point distance z away from the middle point of the two wires.


    2. Relevant equations

    E(left wire) = [tex]\lambda[/tex].K .integration (1/r^2). dr
    E(right wire) = - [tex]\lambda[/tex].K .integration (1/r^2). dr


    3. The attempt at a solution
    r = (x^2 + d^2 + z^2)
    dr = dx
    i don't know what should be the limit of integration and i don't know if whatever i did is right or not.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 26, 2008 #2
    hey, I have attached a figure. I thought this might make your life a little bit easier...i think you have to calculate potential and then calculate E, otherwise it's gonna be hard. I think the x and z components cancel each other. so, you will only have y component of electric field. good luck.
     

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  4. Oct 26, 2008 #3

    gabbagabbahey

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    Do you know the solution for a single infinite wire carrying charge density [itex]\lambda[/itex]? If so, you can use the superposition principle and avoid having to carry out any integration.
     
  5. Oct 26, 2008 #4

    E&M

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    i know the solution for single infinite wire at a point z distance above the center of the wire but since, here we also have to consider y direction, I am not sure how to do it.
     
  6. Oct 26, 2008 #5

    gabbagabbahey

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    Well, what is the solution "for single infinite wire at a point z distance above the center of the wire"?
     
  7. Oct 26, 2008 #6

    E&M

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    K. 2 (lambda) / z for the wire with charge density = positive. lambda
     
  8. Oct 26, 2008 #7

    gabbagabbahey

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    Good, so for that solution z is the distance from the wire....What is the distance from the point (0,0,z) to either of the two wires in the new problem?
     
  9. Oct 26, 2008 #8

    E&M

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    it is (x^2 + y^2 + z^2) ^.5
     
  10. Oct 26, 2008 #9

    gabbagabbahey

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    The distance would be the hypotenuse in the following diagram...
    k
    |
    |
    |\
    | \
    |z \
    |___\_______>j
    d

    wouldn't it?
     
  11. Oct 26, 2008 #10

    E&M

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    if u were asking distance from the center of the wire then it's gonna be (y^2 + z^2) ^.5 where y = d right?
     
  12. Oct 26, 2008 #11

    E&M

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    so, is the E simply going to be K. 2 (lambda) / (d^2 + z^2)^.5 along the direction of (d^2 + z^2) and then when we consider two wires, z components are going to cancel and we calculate and add the y components?
     
  13. Oct 26, 2008 #12

    E&M

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    am I anywhere close?
     
  14. Oct 26, 2008 #13

    gabbagabbahey

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    Yes, the magnitude of E for the wire with +lambda is just K. 2 (lambda) / (d^2 + z^2)^.5. And the magnitude of the other wire is (-)K. 2 (lambda) / (d^2 + z^2)^.5. What are the directions of each of those fields?
     
  15. Oct 26, 2008 #14

    E&M

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    for the individual wires, their x components cancel and they only have y and z components, but if we consider both of the wires, their z components cancel as well and then there will be only y component left pointing towards the wire with -lambda charge density.
     
  16. Oct 26, 2008 #15

    E&M

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    am I right?
     
  17. Oct 26, 2008 #16

    gabbagabbahey

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    Yes, so what is the y-component of E for each wire? (Remember that E points along the hypotenuse in the above diagram)
     
  18. Oct 26, 2008 #17

    E&M

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    individual y component is going to be K* 2 * (lambda) * cos( angle between hypotenuse and d) / (d^2 + z^2)^.5.
    = K* 2 * (lambda) * d / (d^2 + z^2)
    Net y component = 2* K* 2 * (lambda) * d / (d^2 + z^2)
    ????????????
     
  19. Oct 26, 2008 #18

    gabbagabbahey

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    Looks good to me:approve:

    [itex]\vec{E}(z)=\frac{4k \lambda d}{z^2+d^2} \hat{y}[/itex]
     
  20. Oct 26, 2008 #19

    E&M

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    is lambda missing in your equation ?
     
  21. Oct 26, 2008 #20

    E&M

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    thanks a lot
     
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